Subfields joining an algebraic element to another

Question

Let $\alpha$ and $\beta$ be two algebraic numbers over $\mathbb Q$. Say that a subfield $\mathbb K$ of $\mathbb C$ joins $\alpha$ to $\beta$ iff $\beta \in {\mathbb K}[\alpha]$ but $\beta \not\in {\mathbb K}$. Now, if $\mathbb K$ joins $\alpha$ to $\beta$ and we add a completely unrelated algebraic number to $\mathbb K$, we still have a join from $\alpha$ to $\beta$. So it is natural to consider the minimal joins from $\alpha$ to $\beta$, i.e. the joins that are minimal with respect to field inclusion. Let ${\cal M}(\alpha,\beta)$ denote the set of all minimal joins from $\alpha$ to $\beta$. My guesses are that :

1) Any field in ${\cal M}(\alpha,\beta)$ is always contained in the normal (Galois) closure of ${\mathbb Q}(\alpha,\beta)$
2) ${\cal M}(\alpha,\beta)$ is always finite
3) The two facts above should be provable using Galois theory.
Note that 2) follows from 1).

Can anyone confirm this ?

A simple example : ${\cal M}(\sqrt{2},\sqrt{3})$ consists of ${\mathbb Q}(\sqrt{6})$. Indeed, suppose $\mathbb K$ joins $\sqrt{2}$ to $\sqrt{3}$ and $x$ and $y$ are numbers in $\mathbb K$ such that $x+y\sqrt{2}=\sqrt{3}$. If $x \neq 0$ then $\sqrt{3}=\frac{3+x^2-2y^2}{2x} \in \mathbb K$ which is absurd. So $x=0$ and $y=\frac{\sqrt{6}}{2}$.

Answer 1

Finally my guesses seem to be correct : let $d$ be the degree of the extension ${\mathbb K}(\alpha):{\mathbb K} $. There are exactly $d$ field homomorphisms ${\mathbb K}(\alpha)\to \mathbb C$ that coincide with the identity on $\mathbb K$. Each one of this homomorphisms may be extended to a field homomorphism from ${\mathbb K}(\alpha,\beta)$ to $\mathbb C$ (in a generally nonunique way). Let us denote by $H$ the $d$-element set of all the homomorphisms thus obtained.

As darij said, there is a polynomial $P\in {\mathbb K}[X]$ such that $P(\alpha)=\beta$. We may take $P$ of degree smaller than $d$. If we denote the coefficients of $P$ by $p_0,p_1, \ldots , p_{d-1}$, then we have
$\sum_{k=0}^{d-1}p_k \alpha^k=\beta$.
Now, applying the elements of $H$ to this equalities yields
$\sum_{k=0}^{d-1}p_k h(\alpha)^k=h(\beta)$
for any $h\in H$. We see now that the $p_k$ may be retrieved as solutions of a $d\times d$ system all of whose coefficients are in the closure of ${\mathbb Q}(\alpha,\beta)$. The determinant of this system is a Van der Monde determinant on the distinct conjugates of $\alpha$, so it's nonzero.

This shows statement 1) (and 2)). Although darij's proof is wrong as shown in the comments, it may still be that his statement a) is correct (so that "Galois closure" may be supressed in the statement of 1)). Pheraps one could show it the Galois way, by showing that any field homomorphism that coincides with the identity on ${\mathbb Q}(\alpha,\beta)$ in fact preserves the coefficients of $P$.

Answer 2

EDIT: Okay, wrong. Sorry for spamming.

This looks just too easy, so I guess I'm doing something stupid.

I'll prove that

(a) whenever a subfield $K$ of $\mathbb{C}$ satisfies $\beta\in K\left[\alpha\right]$, then $\beta\in\left(K\cap\mathbb{Q}\left(\alpha,\beta\right)\right)\left[\alpha\right]$.

Adding to this the trivial observation that

(b) whenever a subfield $K$ of $\mathbb{C}$ satisfies $\beta\not\in K$, then $\beta\not\in K\cap\mathbb{Q}\left(\alpha,\beta\right)$,

we see that whenever a subfield $K$ of $\mathbb{C}$ joins $\alpha$ to $\beta$, its subfield $K\cap\mathbb{Q}\left(\alpha,\beta\right)$ does the same. This obviously settles 1) (even without the normal closure) and therefore 2).

So let's prove (a) now: Since $\beta\in K\left[\alpha\right]$, there exists a polynomial $P\in K\left[X\right]$ such that $\beta=P\left(\alpha\right)$. Since $K\cap\mathbb{Q}\left(\alpha,\beta\right)$ is a vector subspace of $K$ (where "vector space" means $\mathbb{Q}$-vector space), there exists a linear map $\phi:K\to K\cap\mathbb{Q}\left(\alpha,\beta\right)$ such that $\phi\left(v\right)=v$ for every $v\in K\cap\mathbb{Q}\left(\alpha,\beta\right)$ (this may require the axiom of choice for infinite $K$, but if you want to consider infinite field extensions of $\mathbb{Q}$ I think you can't help but use the axiom of choice). Applying the linear map $\phi$ to every coefficient of the polynomial $P\in K\left[X\right]$, we get another polynomial $Q\in \left(K\cap\mathbb{Q}\left(\alpha,\beta\right)\right)\left[X\right]$ which also satisfies $\beta=Q\left(\alpha\right)$ (since all powers of $\alpha$ and $\beta$ lie in $K\cap\mathbb{Q}\left(\alpha,\beta\right)$ and thus are invariant under $\phi$). Thus, $\beta\in\left(K\cap\mathbb{Q}\left(\alpha,\beta\right)\right)\left[\alpha\right]$, qed.

Can anyone check this for nonsense?